Question

If $y=\sqrt{\frac{1-x}{1+x}}$ then find $(1-x^2)\frac{dy}{dx}+y$ =

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

$y=\sqrt{\frac{1-x}{1+x}}$

$y^2=\frac{1-x}{1+x}$

$y.y=\frac{(1+x)(-1)-(1-x)\left(1\right)}{(1+x{)}^2}$

$y.y=\frac{-1-x-1+x}{(1+x{)}^2}$

$yy=\frac{-2}{(1+x{)}^2}$

$y\frac{dy}{dx}=\frac{-2}{(1+x{)}^2}$

$\frac{dy}{dx}=\frac{-2}{y(1+x{)}^2}$

Given

$(1-x^2)\left(\frac{dy}{dx}\right)+y$

$=(1-x^2)\frac{-2}{y(1+x{)}^2}+y$

$=\frac{-(1-x)}{(1+x)}+y$

$=(1-\frac{(1-x)}{(1+x)}.\frac{1}{y^2})$

$=y(1-\frac{(1-x)}{(1+x)}\times \frac{1+x}{1-x})$

$=0$